|
Answer» Hey there !!
➡ Given :-
→ A ∆ABC in which AD and CE are medians and  B = 90° .
➡ To prove :-
→ 5AC² = 4( AD² + CE² ) .
➡ Proof :-
In ∆ABC, B = 90° .
•°• AC² = AB² + BC² .......(1) .
[ By Pythagoras theorem ] .
In ∆ABD, B = 90° .
•°• AD² = BD² + AB² .
[ By Pythagoras theorem ] .
=> AD² = ( ½BC )² + AB² .
=> AD² = ¼BC² + AB² .
=> 4AD² = BC² + 4AB² ..............(2).
In ∆BEC, B = 90° .
•°• CE² = BE² + BC² .
=> CE² = ( ½AB )² + BC² .
=> CE² = ¼AB² + BC² .
=> 4CE² = AB² + 4BC²................(3) .
▶ On adding equation (2) and (3), we get
=> 4AD² + 4CE² = BC² + 4AB² + AB² + 4BC² .
=> 4( AD² + CE² ) = 5BC² + 5AB² .
=> 5( BC² + CE² ) = 4( AD² + CE² ) [ Using (1) ] .
•°• 5AC² = 4( AD² + CE² ) .
✔✔ HENCE , it is proved ✅✅.
THANKS
#BeBrainly.
|
